Calculating Focal Length for Real and Inverted Image | Simple Optics

[mikehibbert]

[SOLVED] Simple Optics - being stupid?!

Homework Statement

There's an object 1.5m from a white screen. What is the focal length of a lens needed to form a real and inverted image on the screen with magnification of 4.0?
How far from the object should the lens be placed?

Homework Equations

-M=S'/S

1/f=1/S'+1/S

The Attempt at a Solution

My problem is the first part of the question, what to use for S and S', surely there are an infinite number of possibilities?

 

[mgb_phys]

You have a couple of extra constraints.
1, the lens to image_distance is the focal length
2, magnification = f / (f - object_distance)
3, object_distance + image_distance = 1.5m

 

[mikehibbert]

I don't see how point 1 helps me sorry?

I understand that objectdistance + imagedistance = 1.5m

but that gives the distances sooo many possibilities?

 

[mgb_phys]

image distance i = f, object distante = o
i + o = 1.5m
o = 1.5m - i = 1.5m -f

m = f / ( f - o ) = 4

subsitute o and solve for f

 

[mikehibbert]

right,

if i follow that through i get f=8/3

but then putting f back into m=f/(f-o) to find o,

i get o=2, which can't be right because the object is only 1.5m from the screen?

 

[mgb_phys]

Sorry my fault, image distance is only f if the object is at infinity (i'm an astronomer - objects are always at infinity!) I meant to say:
1/f = 1/o + 1/i and o+i=1.5
m = f / (f - o) = 4

Rearrangement is a little trickier

 

[mikehibbert]

i'm really struggling with this... i can't seem to get f out of the maths?!

 

[kdv]

Homework Statement

There's an object 1.5m from a white screen. What is the focal length of a lens needed to form a real and inverted image on the screen with magnification of 4.0?
How far from the object should the lens be placed?

Homework Equations

-M=S'/S

1/f=1/S'+1/S

The Attempt at a Solution

My problem is the first part of the question, what to use for S and S', surely there are an infinite number of possibilities?

maybe I am missing something here but S'/S = -M = 4 so S'= 4S, right? And S+S' = 1.5 meter so a simple substitution gives you both S and S'!

Then you may find the focal length using 1/f = 1/S + 1/S'. Iam probably missing something because it seems straightforward to me.

 

[mikehibbert]

You're right... i feel like SUCH an idiot haha.

Thanks!